已知f(x)除以x-1,x^2-2x+3的余式分别为2,4x+6,则f(x)除以(x-1)(x^2-2x+3)的余式为

设f(x)除以(x-1)(x^2-2x+3)余式为A,则有
f(x)=k(x-1)+2
f(x)=m(x^2-2x+3)+4x+6
f(x)=n(x-1)(x^2-2x+3)+A (1)
用(x-1)去除(1)式,可得 f(x)/(x-1)=n(x^2-2x+3)+A/(x-1)=[n(x^2-2x+3)+a]+2/(x-1)
用(x^2-2x+3)去除(1)式,可得 f(x)/(x^2-2x+3)=n(x-1)+A/(x^2-2x+3)=[n(x-1)+b]+(4x+6)/(x^2-2x+3)
可得 A=a(x-1)+2=b(x^2-2x+3)+4x+6
令a=bx+c,得 (bx+c)(x-1)=bx^2-(b-c)x-c=bx^2-(2b-4)x+3b+4
两端比较得 b-c=2b-4,-c=3b+4；解得b=-4,c=8
∴ a=bx+c=-4x+8=-4(x-1)+4,b=-4
∴A=a(x-1)+2=-4(x-1)^2+4(x-1)+2=-4(x^2-2x+3)+4x+6=-4x^2+12x-6
即f(x)除以(x-1)(x^2-2x+3)的余式为 -4x^2+12x-6